91Ó°ÊÓ

The given power series is $\underset{k=0}{\overset{∞}{∑}}\frac{{\left(-1\right)}^k}{2k+1}{x}^{2k+1}$

Let us assume $b_k=\frac{{\left(-1\right)}^k}{2k+1}{x}^{2k+1}$and

$$\begin{gathered} b_{k+1}=\frac{{\left(-1\right)}^{k+1}}{2\left(k+1\right)+1}{x}^{2\left(k+1\right)+1} \\ {b}_{k+1}=\frac{{\left(-1\right)}^{k+1}}{2k+3}{x}^{2k+3} \end{gathered}$$

Ratio for the absolute convergence is

$$\begin{gathered} \underset{k→∞}{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k→∞}{\lim }\left|\frac{{\displaystyle \frac{{\left(-1\right)}^{k+1}}{2k+3}}{x}^{2k+3}}{{\displaystyle \frac{{\left(-1\right)}^k}{2k+1}}{x}^{2k+1}}\right| \\ =\underset{k→∞}{\lim }\left|\frac{-1\left(2k+1\right)}{2k+3}{x}^2\right| \\ =\underset{k→∞}{\lim }\left|x^2\right|\frac{2k+1}{2k+3} \end{gathered}$$

Here the limit is $\left|x^2\right|$. So, by the ratio test of absolute convergence, we know that series will converge absolutely when$\left|x^2\right|<1$that is$-1<x<1$.

Now, since the intervals are finite so we analyse the behavior of the series at the endpoints.

So, when$x=-1$

$$\begin{gathered} \underset{k=0}{\overset{∞}{∑}}\frac{{\left(-1\right)}^k}{2k+1}{x}^{2k+1}=\underset{k=0}{\overset{∞}{∑}}\frac{{\left(-1\right)}^k{\left(-1\right)}^{2k+1}}{2k+1} \\ =\underset{k=0}{\overset{∞}{∑}}\frac{{\left(-1\right)}^{3k+1}}{2k+1} \end{gathered}$$

The result is the alternating multiple of the harmonic series, which converges conditionally.

So, when$x=1$

$$\begin{gathered} \underset{k=0}{\overset{∞}{∑}}\frac{{\left(-1\right)}^k}{2k+1}{x}^{2k+1}=\underset{k=0}{\overset{∞}{∑}}\frac{{\left(-1\right)}^k{\left(1\right)}^{2k+1}}{2k+1} \\ =\underset{k=0}{\overset{∞}{∑}}\frac{{\left(-1\right)}^k}{2k+1} \end{gathered}$$

Again, the result is the alternating multiple of the harmonic series, which converges conditionally.

Therefore, the interval of convergence of the power series is$\left[-1,1\right]$